Answer:
-0.94492 atm
Explanation:
[tex]P_1[/tex] = Initial pressure = 1 atm
[tex]V_1[/tex] = Initial volume
[tex]T_1[/tex] = Inital temperature = 9°C
[tex]P_2[/tex] = Final pressure
[tex]V_2[/tex] = Final volume = [tex]15V_1[/tex]
[tex]T_1[/tex] = Final temperature = -40°C
We have the relation
[tex]\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2=\dfrac{P_1V_1T_2}{T_1V_2}\\\Rightarrow P_2=\dfrac{1\times V_1\times (273.15-40)}{(273.15+9)\times 15V_1}\\\Rightarrow P_2=0.05508\ atm[/tex]
Gauge pressure is given by
[tex]P_g=P_2-P_a\\\Rightarrow P_g=0.05508-1\\\Rightarrow P_g=-0.94492\ atm[/tex]
The gauge pressure is -0.94492 atm